Raising the bar

The Lagrangian multiplier interpretation

Motivation. Consider the problem of utility maximization $latex \max u(x,y)$ under the budget constraint $latex p_xx+p_yy=M.$ Suppose the budget $latex M$ changes a little bit and let us see how the solution of the problem depends on changes in the budget. The maximized value of the utility will depend on $latex M$, so let us reflect this dependence in the notation $latex U(M)=u(x^\ast(M),y^\ast(M))$ where, for each $latex M$, $latex (x^\ast(M),y^\ast(M))$ is the maximizing bundle. The value of the Lagrangian multiplier $latex \lambda$ in the FOC's $latex \frac{\partial L}{\partial x}=0$, $latex \frac{\partial L}{\partial y}=0$, $latex \frac{\partial L}{\partial \lambda }=0$ will also depend on $latex M$: $latex \lambda =\lambda(M).$ The property that we will prove is

$latex \frac{dU}{dM}=\lambda (M),$

that is, the Lagrangian multiplier measures the sensitivity of the maximized utility function to changes in the budget.

General formulation

The main problem is to maximize $latex f(x,y)$ subject to $latex g(x,y)=0$. However, instead of a fixed constraint we consider a set of constraints perturbed by a constant $latex c$: $latex g(x,y)+c=0.$ Here $latex c$ varies in a small neighborhood $latex (-\varepsilon ,\varepsilon)$ of zero (which corresponds to varying $latex M$ in a neighborhood of some $latex M_{0}$ in the motivating example). Now everything depends on $latex c$, and differentiation with respect to $latex c$ will give the desired result.

Employing the constraint

As before, we assume the implicit function existence condition:

$latex \left( \frac{\partial g}{\partial x},\frac{\partial g}{\partial y}\right)\neq 0.$

Changing the notation, if necessary, we can think that it is the last component of the gradient that is not zero: $latex \frac{\partial g}{\partial y}\neq 0.$ This condition applies to the perturbed constraint too and guarantees existence of the implicit function, which now depends also on $latex c$: $latex y=y(x,c)$. Plugging it into the constraint we get $latex g(x,y(x,c))+c=0$, and differentiation yields

(1) $latex \frac{\partial g}{\partial y}\frac{\partial y}{\partial c}+1=0.$

Employing the FOC's

Critical assumption. For each $latex c\in (-\varepsilon ,\varepsilon )$ the perturbed maximization problem has a solution. At least in our motivating example, this assumption is satisfied.

The maximized objective function is denoted $latex F(c)=f(x,y(x,c)).$ At each $latex c,$ we have the right to use the FOC's for the Lagrangian. One of the FOC's is

$latex \frac{\partial f}{\partial y}+\lambda (c)\frac{\partial g}{\partial y}=0.$

To make use of (1), multiply this by $latex \frac{\partial y}{\partial c}$:

$latex 0=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}+\lambda(c)\frac{\partial g}{\partial y}\frac{\partial y}{\partial c}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}-\lambda (c).$

It follows that

$latex \frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda (c).$

With $latex c=0$ we have $latex \lambda(c)=\lambda$, the Lagrange multiplier for the unperturbed problem, and as a result

$latex \frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda.$